Let $\alpha$ belong to $S_n$. Prove that $|\alpha|$ divides $n!$

Question

From Gallian's "Contemporary Abstract Algebra", Part 2 Chapter 5

It looks like using Lagrange's theorem would work, since $|S_n| = n!$ and $\langle\alpha\rangle$ is a subgroup of $S_n$. However, that hasn't been covered in the book at this point, so I'm assuming a different solution is expected

$\alpha$ can be broken up into disjoint cycles $\alpha_1\dots\alpha_m$ such that $|\alpha_1| + \dots +|\alpha_m| = n$, and then $|\alpha| = \operatorname{lcm}(|\alpha_1|, \dots, |\alpha_n|)$. Don't know how to continue though

Answer 1

You've almost solved it yourself! The numbers $|\alpha_i|$ are all between $1$ and $n$ so, from what you wrote, $|\alpha|$ is the least common multiple of a set of numbers between $1$ and $n$. On the other hand $n!$ is the product of all the numbers from $1$ to $n$. One definition of the least common multiple of a set of numbers is that it divides any number which is divisible by all the numbers in the set.

Answer 2

Hint

The order of any cycle of size $k$ is $k$, and thus less than or equal to $n$. And thus divides $n!$.

Then there's the universal property of the lcm: $a,b\mid c\implies \rm{lcm}(a,b)\mid c$.

Thus we go around (or under) Lagrange.